V = IR, so if you double the voltage without changing the resistance, the current will also double.
The current will increase by a factor of 6 times, and the light bulb will be brighter, and may burn out quickly.
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Current will go up by a factor of 6 times in that scenario.
The battery cannot be removed quickly! Inductance of the wiring stores energy and keeps the current flowing and decaying smoothly. There will be an opposite polarity, higher voltage pulse on the battery terminals, limited by the circuit capacitance only.
The battery cannot be removed quickly! Inductance of the wiring stores energy and keeps the current flowing and decaying smoothly. There will be an opposite polarity, higher voltage pulse on the battery terminals, limited by the circuit capacitance only.
Current will go up by a factor of 6 times in that scenario.
V = IR, so if you double the voltage without changing the resistance, the current will also double.
It wil be on
Voltage, current, and resistance can be related by this formula. V = iR, where I is the current. Assuming that the voltage stays constant, current will decrease. Hope this helps!
it can exchange the current is much larger and has Avery low resistance when it is turned on.the mosfet is not removed from the circuit when the supply is on because it flow the high current.
The computer only resets when the battery is removed. The alternator just charges the battery, so if the battery is left connected when the alternator is replaced, the computer should not reset.
Nothing will happen to circuit..... as usual the circuit would be supplying 220v(if india) and certain current...but there is no bulb to consume power...
Opening any circuit will stop the current from flowing.
assume the following configuration: battery connected to 2 parallel resistors with an ammeter in series with the battery... observe the current measurement ... remove one of the resistors .... observe the current again, it will have decreased: if the resistors were of equal value, the current will decrease to half of its original value when one of the resistors is removed. Mathematics: V=IR (V- voltage, I - current, R - resistance in a parallel circuit, R=(R1*R2)/(R1+R2) where R1 and R2 are the values of resistance of the resistors. Before removal- Ib=V*(R1+R2)/(R1*R2) After removal (assume R2 is removed)- Ia=V/R1 so Ia/Ib=(R1*R2)/(R1*(R1+R2)) or Ia=Ib*(R2/(R1+R2) if R1=R2 then Ia=Ib*R2/(2*R2) or Ia=Ib/2 so the current after is 1/2 of that before.